Answer:
[tex]a.\ \hat p=0.2877\\\\b.\ ME=0.0263\\\\c.\ 0.2614<\hat p<0.3140[/tex]
d. -Sample selection was random
-Individual observations are independent of each other
np≥10
Step-by-step explanation:
a. The point estimate of a sample proportion is obtained using the formula;
[tex]\hat p=\frac{x}{n}\\\\x-sample \ size\\n-population\ size\\\hat p-point \ estimate\\\\\\\\\therefore \hat p=\frac{328}{1140}\\\\\\=0.2877[/tex]
Hence, the point estimate of the proportion of the population is 0.2877
b. The desired margin of error is the calculated using the point estimate value as follows:
[tex]ME=z\sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\z_{0.025}=1.96\\\\\hat p=0.2877\\\\\therefore ME=1.96\times \sqrt{\frac{0.2877(1-0.2877)}{1140}} \\\\=0.0263[/tex]
Hence, the desired margin of error for the sample proportion is 0.0263
c. Given a confidence level of 95%, the confidence interval can be calculated as:
[tex]CI_{95\%}=\hat p\pm ME\\\\=0.2877\pm 0.0263\\\\=[0.2614,0.3140][/tex]
Hence, the confidence interval at 95% confidence level is 0.2614<p<0.3140
#We are 95% confident that the interval estimate contains the desired proportion.
d. The assumptions are:
-The sample size is is more than 10 or equal to 10:
[tex]n\hat p\geq 10\\n\hat p=0.2877\times 328=94.37\geq 10\\\\\therefore np\geq 10[/tex]
-The selection was from a randomized experiment.
-The individual observations were independent of each other.
